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## anonymous 4 years ago Solve the non-linear inequality. Express the solution in interval notation. x^4 is greater than x^2

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1. anonymous

$x^4>x^2$

2. anonymous

I take the square root of both sides and get the plus or minus square root of x^2 is greater than x, then I take the square root of x^2 and x to get x is greater than the plus or minus square root of x

3. anonymous

$x ^{4}>x ^{2}$ $x ^{4}-x ^{2}>0$ $x ^{2}(x ^{2}-1)>0$ $x ^{2}>0 or x ^{2}>1$ x>0 or x<-1 or x>1 then x<-1 and x>0 will be the interval...

4. klimenkov

$$x^2(x^2-1)>0$$ divide it by $$x^2$$: $$(x-1)(x+1)>0$$ $$x\in(-\infty;-1)\bigcup(1;\infty)$$

5. anonymous

why is it greater than 0

6. anonymous

oh subtract the x^2?

7. anonymous

yep//....

8. anonymous

@znimon understood the process?

9. anonymous

almost, what about after x^2(x^2-1) is greater than 0

10. klimenkov

$$x^2$$ is always bigger than 0. It equals 0 when $$x=0$$. But it doesn't satisfy the inequality.

11. anonymous

took x^2 common in both the terms....

12. anonymous

yep got that

13. anonymous

nops.. while splitting the inequality x has to be >0

14. anonymous

will you break it down after the x^2(x^2-1) is greater than 0

15. anonymous

yep.... both the split terms are compared to the inequalities...

16. anonymous

I understand how it works so I need to find all the solutions for x that satisfy the inequality

17. anonymous

thanks now I'll try some on my own. Will you recommend a problem that has the same properties?

18. anonymous

@sriramkumar

19. anonymous

try this one... $x ^{8}-x ^{2}<0$ remeber x^3 has different properties when compared with x^2 .... have a good time!!!!1

20. anonymous

thanks

21. anonymous

@znimon you're welcome :)))))))))))))))))

22. anonymous

$\ \ x^4\ge x^2\\ \ \ x^2[x^2]\ge x^2\\x^2\text{ is either 0 non-zero.}\\ \text{if }x^2\text{ is 0,}\\ \ \ 0\times0=0\\ \ \ 0 = 0\\ \text{therefore }x^2[x^2]\ge x^2\\ \text{otherwise,}\\ \ \ x^2\ge1\\ \ \ |x|\ge\sqrt1$

23. anonymous

is either 0 or non-zero **

24. anonymous

solution is \[x <0, x <1] so (negative infinity, to 0 not including zero)?

25. anonymous

for x^3-x^2 is less than 0

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